There is a cylindrical shape sweet potato with an initial uniform temperature of 25°C and it has an average diameter of 70 mm and length of

Question

There is a cylindrical shape sweet potato with an initial uniform temperature of 25°C and it has an average diameter of 70 mm and length of 150 mm. Thermal conductivity and diffusivity of the sweet potato are 0.62 W/m · °C and 0.2 x 10-6 m2/s, respectively. It is to be cooled by refrigerated air at 1.8°C. The average heat transfer coefficient between the potato and the air is 25 W/m2 · °C. Determine how long it will take for the center temperature of the potato to drop to 5.5°C.

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Khánh Gia 5 years 2021-08-29T09:25:06+00:00 1 Answers 18 views 0

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  1. Verity
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    2021-08-29T09:26:57+00:00
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    Explanation:

    The given data is as follows.

        T_{o} = 25^{o}C,    d = 70 mm = 70 \times 10^{-3} m

        k = 0.62 W/m^{o}C,     \alpha = 0.2 \times 10^{-6} m/sec

        T_{\infinity} = 1.8^{o}C,    v = 4 m/sec

           h = 25 W/m^{2}^{o}C,  T_{c} = 5.5^{o}C

    First, we will calculate the value of B_{i} as follows.

             B_{i} = \frac{h \times d}{4k}

                        = \frac{25 \times 70 \times 10^{-3}}{4 \times 0.62}

                        = 0.7056

    Since, 0.7056 is greater than 0.1.

    As,  \frac{T_{c} - T_{\infinity}}{T_{o} - T_{infinity}} = 0.159

                    = 0.16

    Also here,

                 \frac{\alpha T}{d^{\gamma}} = 1.9

               \frac{0.2 \times 10^{-6} \times T}{\frac{(70 \times 10^{-3}}{4})^{\gamma}} = 1.9

                     T = 46.18 hrs

    Thus, we can conclude that it will take 46.18 hrs for the center temperature of the potato to drop to 5.5^{o}C.

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